Center of Mass Rotational Motion

[xxphysics]

Homework Statement

The system shown in (Figure 1) consists of two balls Aand B connected by a thin rod of negligible mass. Ball Ahas three times the inertia of ball B and the distance between the two balls is ℓ. The system has a translational velocity of v in the x direction and is spinning counterclockwise at an angular speed of ω=2v/ℓ.

Homework Equations

mArA = mBrB
rA + rB = ℓ
ω = 2v/ℓ
vA = v - (2v/ℓ)rA
vB = v + (2v/ℓ)rB

3. The Attempt at a Solution
mArA = mBrB ; mA = 3mB

3mBrA = mBrB
3rA = rB

rA + rB = ℓ
rA + 3rA = ℓ
4rA = ℓ
rA/ℓ =1/4

ω = 2v/ℓ

vA = v(1 - 2rA/ℓ)
= v(1 - 2/4)
= (1/2)v

vB = v(1 + 2rB/ℓ)
= v(1 + 2(3rA)/ℓ)
= v(1 +6rA)/ℓ)
= v(1 + 6/4)
= (3/2)v

vA/vB = [(1/2)v]/[(3/2)v] = (1/2)/(3/2) = (1/2)x(2/3) = 1/3

1/3 is not the correct answer, where am I going wrong?

 

[TSny]

Homework Statement

The system shown in (Figure 1) consists of two balls Aand B connected by a thin rod of negligible mass. Ball Ahas three times the inertia of ball B and the distance between the two balls is ℓ. The system has a translational velocity of v in the x direction and is spinning counterclockwise at an angular speed of ω=2v/ℓ.

The problem statement doesn't state what you are to find.

vB = v(1 + 2rB/ℓ)
= v(1 + 2(3rA)/ℓ)
= v(1 +6rA)/ℓ)
= v(1 + 6/4)
= (3/2)v

EDIT:

vA = v - (2v/ℓ)rA
vB = v + (2v/ℓ)rB

Do you have the correct signs here?

 

[xxphysics]

The problem statement doesn't state what you are to find.
EDIT:

Do you have the correct signs here?

Oh, I'm sorry. I'm supposed to find the ration of Va/Vb. Ans for the second portion, should it be 10/4v then for vB? So 1/5 is the ratio ?

 

[xxphysics]

Oh, I'm sorry. I'm supposed to find the ration of Va/Vb. Ans for the second portion, should it be 10/4v then for vB? So 1/5 is the ratio ?

Also, yes I believe I have the correct signs.

 

[TSny]

Does V_A represent the speed of A relative to the "earth"?

In what direction is the velocity of A due to the translational motion?

In what direction is the velocity of A due to the rotational motion?

How would these combine (add or subtract)?

 

[xxphysics]

Does V_A represent the speed of A relative to the "earth"?

In what direction is the velocity of A due to the translational motion?

In what direction is the velocity of A due to the rotational motion?

How would these combine (add or subtract)?

Ohhhhh, thank you! So the answer should be -3, or 3 since its a ratio.

 

[haruspex]

Ohhhhh, thank you! So the answer should be -3, or 3 since its a ratio.

That's not what I get. Please post your most recent working.

 

[xxphysics]

That's not what I get. Please post your most recent working.

vA = v(1 - 2rA/ℓ)
= v(1 + 2/4)
= (3/2)v

vB = v(1 + 2rB/ℓ)
= v(1 + 2(3rA)/ℓ)
= v(1 +6rA)/ℓ)
= v(1 - 6/4)
= (-1/2)v

vA/vB = [(3/2)v]/[(-1/2)v] = (3/2)/(-1/2) = (3/2)x(-2/1) = -3 but its a ratio so just 3. It said it was correct ?

 

[haruspex]

vA = v(1 - 2rA/ℓ)

As TSny indicated, the sign above looks wrong, but maybe you are defining r_A in such anway that its value will be negative.

= v(1 + 2/4)

You seem to have substituted r_A=-l/4. How did you get that?

 

[xxphysics]

You seem to have substituted rA=-l/4. How did you get that?

mArA = mBrB ; mA = 3mB

3mBrA = mBrB
3rA = rB

rA + rB = ℓ
rA + 3rA = ℓ
4rA = ℓ
rA/ℓ =1/4I got the answer of 3 which was correct. Thank you

 

[haruspex]

mArA = mBrB ; mA = 3mB

3mBrA = mBrB
3rA = rB

rA + rB = ℓ
rA + 3rA = ℓ
4rA = ℓ
rA/ℓ =1/4

Sure, but that's not -l/4.

Anyway, I made a mistake. I agree with your 3:1, but your sign errors were confusing.